Pshs Upcat Review Bio (Part 2) Answer Guide

Genetics

Molecular Dogma

Chemical Biology

Cell Division

Note: The discussion here is a very general overview of topics

that MIGHT come out in the UPCAT. Your discussion of these

topics in Bio3 will definitely be more formal and more

detailed. Also, the more complicated problems here will most

probably NOT come out in the UPCAT, but the basic topics

from which they stem might.

What determines the sex of an unborn child?

The number of days between ovulation and fertilization

The chromosome content of the sperm

The chromosome content of the ovum

Diet of the father before sexual intercourse

G1

• 2 varieties of sex chromosomes in humans: X & Y

– XX = female, XY = male

• The two sex chromosomes segregate during meiosis, and each

gamete receives one

– In the ovaries: In the testes:

• At the moment of conception:

X

X

X

X

X

Y

X

Y

X XX

XX Y

X

Y

Female Male

In pea plants, spherical seeds (S) are dominant to

dented seeds (s). In a genetic cross of two plants that

are heterozygous for the seed shape, what fraction of

the offspring would be spherical seeds?

1

1/2

1/4

3/4

G2





Alleles (alternative form) of the gene

(segment of DNA) for seed shape

• Dominant: trait is expressed all the

time

o SS or Ss spherical

• Recessive: trait is expressed only if the

organism has 2 copies of the allele

o ss dented

• Heterozygous: has one of

each kind of allele (Ss)

• Homozygous: has two of the

same kind of allele

o Homozygous dominant: SS

o Homozygous recessive: ss




the offspring would be spherical seeds?the parents both have the genotype (genetic makeup) Ss and

are both spherical (the phenotype, or physical and observable

expression of the genotype)

during meiosis, the alleles for seed shape are segregated as

follows, for both male and female parents:

S

s

S

s





• During fertilization, the possible combinations of gametes

displayed in the table below, called a Punnett square:

S s

S SS Ss

s Ss ss

Gametes from parent 1

Gametes from parent 2

Possible genotypes of offspring





• The resulting ratios of genotypes and phenotypes are:

• Take note that these are always the resulting ratios of a

monohybrid cross

Genotype Ratio Phenotype Ratio

SS 1/4Spherical seed (SS or Ss) 3/4

Ss 2/4

ss 1/4 Dented seed 1/4

4/4 4/4

mono: dealing w/ a single trait

hybrid: parents are heterozygous

A man with type A blood marries a woman with type

B blood. Their child has type O blood. The genotypes

of the father and the mother, respectively, are

AO; BB

AO; BO

AA; BB

AB; BB

G3

• The ABO blood type is controlled by a single gene with three

alleles: IA, IB, and i*. IA and IB are codominant; IA and IB are

both dominant over i.

• The following are the genotypes of the different ABO blood

group phenotypes:

* For simplicity, IA is sometimes represented as A, IB as B, and i

as O

Phenotype Genotype

Type A IAIA or IAi

Type B IBIB or IBi

Type AB IAIB

Type O ii

A man with type A blood marries a woman with type

B blood. Their child has type O blood. The genotypes

of the father and the mother, respectively, are…

You can construct a Punnett square using the

different combinations of parental genotypes to see

which combination will yield a type O child.

Individual Phenotype Genotype

Father Type A IAIA or IAi

Mother Type B IBIB or IBi

Child Type O ii

• (Additional info) Immunological aspects of the ABO blood

group:

Recipient phenotype Donor phenotype(s)

Type A

• has A antigens

• produces antibodies vs. B antigens

A or O

Type B

• has B antigens

• produces antibodies vs. A antigens

B or O

Type AB (universal recipient)

• has both A and B antigens

• does not produce antibodies to either A or B

antigens (because they have both)

A, B, AB, or O

Type O (universal donor)

• does not have either A or B antigens;

• produces antibodies vs. both A and B antigens

O

The diagram below shows the family tree of a family with

phenylketonuria (PKU). PKU is a disease that is expressed in

homozygous recessive individuals.

Which of the following correctly describes the genotype of

individuals in the family tree?A. P and Q are heterozygous

B. P and Q are homozygous dominant

C. R and S are homozygous dominant

D. R is homozygous dominant, S is heterozygous dominant

G4

• A pedigree is a family tree describing the interrelationships

of parents and children across the generations

• Legend:

– is a ; is a

– indicates a mating, with offspring listed below in

their order of birth

– shaded symbols stand for individuals with the trait

being traced

If we let P = dominant allele of the gene associated with the occurrence of PKU, and

p = recessive allele, this means that all shaded symbols have a genotype of pp.

In determining the unknown genotypes in a pedigree, it is usually helpful to focus

first on matings that produce homozygous recessive offspring, since this can narrow

down your choices for the parents’ genotypes.

pp

pp pp

pp

Let’s focus first on the mating of R and S. Since they are both unaffected, and they

have an affected offspring (which means one or both of them must carry but not

express the recessive allele) their possible genotypes are P_ x P_. By constructing

Punnett squares of the possible genotypes of the parents, we can get these results:

pp

pp pp

pp

Parents Offspring

PP x Pp 2 PP: 2 Pp

Pp x Pp 1 PP: 2 Pp: 1 ppThese results indicate that R and S

are both heterozygous.

pp

pp pp

pp

Parents Offspring

PP x pp All Pp

Pp x pp 2 Pp: 2 pp

At this point, you can already

answer the question, though I

encourage you to try to figure out

the genotype of individual Q.

Next, let’s focus on individual P (unknown genotype) and its mate (genotype pp).

Individual P’s possible genotypes are PP or Pp (can you figure out why?). Again,

constructing Punnett squares for the possible mating combinations would yield the

following results:

Pp Pp

The diagram below shows the steps in protein synthesis.

If the DNA sequence used to produce X is

5’-ATGTGGAAT-3’, what will be the sequence of X?

5'-TACACCTTA-3'

5'-UACACCUUA-3'

3'-TACACCTTA-5'

3'-UACACCUUA-5'

DNA X Protein

MD1

• The flow of genetic information, summarized in the diagram

below, is usually referred to as the central dogma of

molecular biology.

Notice that this diagram is

very much similar to the

diagram in the question.

• DNA (a polymer) carries genetic information via the sequence of its

nucleotides (its monomers).

• Nucleotides are made of

building blocks themselves:

a nitrogenous base,

a pentose (5-carbon) sugar,

and a phosphate group.

• Nitrogenous bases:

– Adenine (A)

– Thymine (T)

(or uracil (U), in RNA)

– Cytosine (C)

– Guanine (G)

Nitrogenous

base

Nucleoside

O

O

O

O P CH2

5’C

3’CPhosphate

group Pentose

sugar

O

• The DNA molecule consists of two nucleotide chains that spiral

around an imaginary axis, forming a double helix.

• The nitrogenous bases of one

chain are linked to those of the

other chain through hydrogen

bonds. The two chains are said

to be complementary. The

base-pairing rules are

as follows:

– adenine with thymine

(or uracil, in RNA)

– cytosine with guanine

• The central dogma illustrates how information in DNA is used to make proteins.

• The nucleotide sequence in DNA (or simply, the DNA sequence) is used as a basis to make an intermediate molecule, RNA. This is the process of transcription.

• The RNA sequence is then used to come up with a sequence of amino acids, which eventually make up a protein. This is the process of translation.

• Going back to the question: If the DNA sequence used to produce X

is 5’-ATGTGGAAT-3’, what will be the sequence of X?

• We are thus being asked to make an RNA sequence given this DNA

sequence.

• Before we start, though, we must know that when an RNA

molecule is made, it elongates from the 5’ end to the 3’ end

(5’3’). Thus its DNA template must be the strand that goes from

3’5’, which means that we need to know the complementary

strand of the DNA sequence given above.

Given strand: 5’-ATGTGGAAT-3’

Complementary strand: 3’-TACACCTTA-5’

This will be the template for the RNA

sequence we are being asked of.

• Going back to the question: If the DNA sequence used to produce X

is 5’-ATGTGGAAT-3’, what will be the sequence of X?

• Now that we know what the sequence of the template DNA is, we

can put together the RNA that will result during transcription:

Template strand: 3’-TACACCTTA-5’

RNA after transcription: 5’-AUGUGGAAU-3’

• Note that the resulting RNA sequence is the same as the DNA

sequence initially given to us, except that the T’s were replaced

by U’s.

• Note, also, that the correct answer is not among the choices given in the problem.

Sorry about that :P

Which component is NOT DIRECTLY involved in the

process known as translation?

ribosomes

RNA

DNA

mRNA

MD2

• Translation is the RNA-directed synthesis of a polypeptide.

• Recall from your lessons on cell structure and function the

functions of ribosomes.

• Given these information, you will then be able to answer

the previous question.

This means that RNA (in particular,

messenger RNA (mRNA), which was

produced by transcription) is the

template for making protein.

Proteins consist of one or more

polypeptides, which are chains of

amino acids.

Based on the genetic code below, which amino acid sequence is

specified by the mRNA sequence UAUCGCACCUCAUAG when the

first triplet (UAU) is used as the start of the reading frame?*

Tyr-Arg-Thr-Ser

Tyr-Arg-Thr-Asp

Tyr-Arg-Thr-Ser

Tyr-Gly-Ser-Leu

* The PDF version of this reviewer

only has tripeptides as choices for A

and B. Please use the choices as

listed here. Sorry about that.

MD3

• The genetic code is made of triplets of nucleotides and the amino acids for which they code.

• During translation, our cells (ribosomes, in particular) “read” the mRNA produced earlier by translation in sets of three nucleotides (codons). As the ribosome encounters a certain codon, the corresponding amino acid is added to the growing polypeptide.

DNA

molecule

Gene 1

Gene 2

Gene 3

DNA strand

(template)

TRANSCRIPTION

mRNA

Protein

TRANSLATION

Amino acid

A C C A A A C C G A G T

U G G U U U G G C U C A

Trp Phe Gly Ser

Codon

3 5

35

• Given the mRNA sequence,

separated into codons:

UAU CGC ACC UCA UAG

• The first codon: UAU

– Locate the 1st nucleotide, U,

on the leftmost column of the

genetic code to the right

– Then locate the 2nd nucleotide,

A, on the top row

– Then locate the 3rd nucleotide,

U, on the rightmost column

– The amino acid found at the

intersection of those three

nucleotides is the amino acid

corresponding to that codon.

• Given the mRNA sequence,

separated into codons:

UAU CGC ACC UCA UAG

• If you work out the rest of the

sequence, you should come up

with: Tyr-Arg-Thr-Ser

• The stop codons – UAA, UGA, and

UAG – are chemical signals to the

cell that indicate the end of

translation

Based on the genetic code below, identify a possible sequence of

nucleotides in the DNA template for an mRNA coding for the

polypeptide sequence Phe-Pro-Lys.

UUU-CCC-AAA

TTT-CCA-AAA

TTC-CCC-AAG

AAA-GGG-UUU

MD4

• Following the previous discussion, we now work our way

back with this problem.

• First, let us list the codons that

code for the given amino acids:

Amino acid Codons

Phe UUU, UUC

Pro CCU, CCC, CCA, CCG

Lys AAA, AAG

• Since we have many choices

for the middle amino acid (aa),

let us focus first on the 1st and

3rd aa’s.

• A Phe-Pro-Lys tripeptide will result from an RNA sequence that

– starts with either 5’-UUU or 5’-UUC

– ends with either AAA-3’ or AAG-3’

• This means that its DNA template would have a sequence such as

- 3’-AAA or 3’-AAG

- TTT-5’ or TTC-5’

Amino acid Codons

Phe UUU, UUC

Pro CCU, CCC, CCA, CCG

Lys AAA, AAG

• With these options for the end sequences:

- 3’-AAA or 3’-AAG

- TTT-5’ or TTC-5’

We can eliminate the obviously wrong answers from the options

given to us. Notice, too, that the eliminated choices below could

have also been discarded at the start because of the uracils in

their sequence (DNA does not have uracil).

A. AAA-GGG-UUU C. TTT-CCA-AAA

B. TTC-CCC-AAG D. UUU-CCC-AAA

• Now that you’re left with just options B and C, you can now look at

the possible sequences for the 2nd aa in order to choose your final

answer. Can you figure out why the BEST answer is C?

Which of the following is NOT true of a codon?

It is the basic unit of the genetic code.

It never codes for more than one amino acid.

It may code for the same amino acid as another codon does.

It consists of three nucleotides.

You can easily answer this by studying the genetic code from the

previous problems.

MD5

Corn plants have been genetically modified to contain a gene

which can produce toxins to kill pests of the plant. Which of the

following arguments is NOT a valid reason for opposing the

widespread use of this genetically modified crop?

The modified chromosomes could cause the chromosomes in the

cell in the body of the consumer to mutate.

The inserted gene could become incorporated into the genome of

weed species growing nearby.

The inserted gene could cause mutation to occur on other parts of

the plant chromosome.

Beneficial insects may inadvertently be killed.

Statements A and C describe possible scenarios. Thus your choices are down

to B or D, both of which have little or no scientific evidence of happening (at

least not yet).

MD6

Some substances were isolated from a chimpanzee that directly

played roles in body structure, immunology, and oxygen

transport. What is the classification of these substances?

nucleic acids

proteins

carbohydrates

lipids

Carbohydrates and lipids are mainly sources of energy, while nucleic

acids store genetic information. Proteins are called the workhorses of

the body, since they perform various functions such as the ones

mentioned in this problem.

CB1

Given the table below, what are substances Y

and Z?

proteins; monosaccharides

fats; fatty acids and glycerol

carbohydrates; amino acids

fats; amino acids

Digestive enzyme Substance digested Resulting simple

molecules

Amylase Polysaccharides Disaccharides

(W) Proteins (X)

Lipase (Y) (Z)

CB2

• The name of the enzyme – lipase – indicates that the enzyme acts

on lipids. Thus the question is: what are the building blocks of

lipids?

• The following table lists the important molecules found in living

organisms, as well as their building blocks. Note that only

carbohydrates, proteins, and nucleic acids are considered polymers

(long molecules consisting of many similar building blocks called

monomers). Lipids consist of two types of smaller molecules.

Biomolecule Building blocks

Carbohydrates Monosaccharides

Proteins Amino acids

Nucleic acids Nucleotides

Lipids 1 glycerol + 3 fatty acids

A cell may use lipids to

hydrolyze food molecules.

provide structural support.

form a barrier between the cell and its environment.

speed up the chemical reaction.

CB3

• Let’s analyze which biomolecules are being referred to by each option:

A. speed up the chemical reaction. these are enzymes, which are proteins

B. form a barrier between the cell and its environment. the barrier refers to the cell membrane, which

consists of a lipid bilayer

C. provide structural support. proteins such as collagen and elastin are known to

provide structural support in animals

D. hydrolyze food molecules.

again, these refer to proteins

Which of the following best describes the difference

between the structure of carbohydrates and fats?

Carbohydrates Fats

A. Do not contain the element

nitrogen

Contain the element nitrogen

B. Ratio of H atoms to O atoms is

2:1

No fixed ratio of H and O

atoms

C. Made of glucose molecules

joined together

Made of fatty acids and

glycerol joined together

D. No fixed ration of H to O

atoms

Ratio of H atoms to O atoms is

2:1

CB4

• All you need to answer the question is to recall the

structure of carbohydrates and fats:

– Carbohydrates: C(H2O)n

– Fats or lipids: 1 glycerol

+ 3 fatty acids

Which of the following is NOT a property of ALL

enzymes?

Enzymes are required in very small quantities.

Temperature and pH affect the rate of enzyme reactions.

Enzymes have active sites of specific shapes.

Enzymes catalyze catabolic reactions only.

You can answer this easily

CB5

Scientists can create new genes and place them in

another organism’s cells. What is this technique?

recombinant DNA technology

cloning

DNA mutation

DNA transfer

You can answer this easily The definitions of these terms are in

Campbell.

CB6

Which is NOT true about the structure of

eukaryotic DNA?

It is composed of phosphates, ribose, and nitrogenous bases.

It contains nucleotides as its basic structural unit.

It is made up of two complementary strands.

It has double helical structure.


CB7

Which of the following statements about

nucleic acids is correct?

All of the above are correct.

The bases and sugar molecules found in DNA and RNA are the

same.

Both DNA and RNA are assembled from nucleoside triphosphates.

In a given sample of DNA the amount of thymine is equal to the

amount of cytosine.

CB8

To evaluate the correctness of each statement, you need to recall a

few facts about nucleic acids.

A. In a given sample of DNA the amount of thymine is equal to the

amount of cytosine.

base-pairing rules: A-T and C-G

B. Both DNA and RNA are assembled from nucleoside triphosphates.

Nucleoside triphosphates are the ingredients of polynucleotide

synthesis

C. The bases and sugar molecules found in DNA and RNA are the

same.

Clue: DNA, RNA

A scientist analyzed several DNA samples to determine the

relative proportions of purine (adenine and guanine) and

pyrimidine (cytosine and thymine) bases. Her data are

summarized in the table below:

Percentages of Nitrogenous Bases in Three Samples

Sample G C A T

I 35 35 15 15

II 40 10 40 20

III 25 25 25 25

Which sample(s) supports the base pairing rules?

A. Sample I only C. Sample I and II

B. Sample II only D. Samples I, II, and III

CB9

• If a sample supports the rules on base pairing (already

mentioned earlier), that would mean that the amounts of G

and C will be equal, and that the amounts of A and T will be

equal.

Note that none of the choices is valid. Again, sorry for that :P

A scientist analyzed several DNA samples to determine the

relative proportions of purine (adenine and guanine) and

pyrimidine (cytosine and thymine) bases. Her data are

summarized in the table below:

Percentages of Nitrogenous Bases in Three Samples

Sample G C A T

I 35 35 15 15

II 40 10 40 20

III 25 25 25 25

If the scientist had analyzed mRNA rather than DNA, what percentage

of uracil would you expect to find in Sample II?

A. 10 C. 35

B. 25 D. 40

For this question, you just need to remember that in RNA uracil is present

rather than thymine.

CB

10

Nerve cells and cardiac (heart) cells have long life spans and do not

undergo mitosis once they are formed. Smooth muscles cells also have

long life spans but are capable of undergoing mitosis. Which of the

following statements is FALSE regarding consequences of injuries to

the spinal cord, heart, or smooth muscle?

All of the above statements are false.

Injuries to the spinal cord, heart, and smooth muscle cause serious

damage and take a very long time to heal.

Injuries to the spinal cord and heart cause serious and permanent

damage.

Damage caused by injuries to smooth muscle may be reversed

because of its capability to regenerate.


CD1

Which of the following stages of mitosis can be

viewed as the opposite of prophase if we consider

only the changes in the nucleus?

telophase

metaphase

interphase

anaphase

CD2

Recall the stages of mitosis:

The following statements regarding mitosis

and meiosis are true EXCEPT

Mitosis results in two genetically identical diploid daughter cells,

while meiosis results in four genetically nonidentical haploid

daughter cells.

The second meiotic division is similar to mitosis in that the

chromosome number is reduced.

Mitosis involves only one round of cell division, while meiosis

involves two rounds.

Replication of DNA occurs before nuclear division begins.

CD3

Recall the stages of meiosis (those of mitosis are on

a previous slide already).

Mature cells no longer divide, so they do not replicate their DNA. A

cell biologist found that there was X amount of DNA in a human nerve

cell. The biologist then measured the amount of DNA in four other

types of human cells; the results are recorded in the chart below.

Complete the chart by filling in the type of cell from these choices:

Cell Amount of DNA Type of Cell

I 2X

II 1.6X

III 0.5X

V X

A. sperm cell

B. bone marrow cell just beginning interphase of the cell cycle

C. skin cell in the S phase of the cell cycle

D. intestinal cell beginning mitosis.

CD4

Here are some clues:

Cell Amount of DNA Type of Cell

I 2X

II 1.6X

III 0.5X

V X Try to figure

this out first.

A. sperm cell

haploid

B. bone marrow cell just beginning interphase of the cell cycle

C. skin cell in the S phase of the cell cycle

D. intestinal cell beginning mitosis.

Choices B, C, and D are all diploid cells. At each of the

stages of the cell cycle mentioned, is there any DNA

replication going on?

“Chance favors the prepared mind.”

- Louis Pasteur

Technology

Pshs Upcat Review Bio (Part 2) Answer Guide